Math Problem Statement

Study the convergence of the given integrals and determine whether each integral is convergent or divergent. If convergent, find its exact value.

Solution

Let’s analyze the problem step by step. The task is to study the convergence of the given integrals:

Part A:
01ln(x)x4/5dx\int_0^1 \frac{\ln(x)}{x^{4/5}} \, dx

Part B:
1ln(x)x4/5dx\int_1^\infty \frac{\ln(x)}{x^{4/5}} \, dx

Analysis for Part A:

  1. Behavior near x=0x = 0:
    The function ln(x)x4/5\frac{\ln(x)}{x^{4/5}} has a potential singularity at x=0x = 0, as ln(x)\ln(x) \to -\infty and x4/5x^{-4/5} \to \infty.
    To study convergence, expand the integral: 01ln(x)x4/5dx\int_0^1 \frac{\ln(x)}{x^{4/5}} \, dx

    Let x=t5x = t^5, so dx=5t4dtdx = 5t^4 \, dt. Under this substitution: ln(x)=ln(t5)=5ln(t)\ln(x) = \ln(t^5) = 5\ln(t) and x4/5=t4x^{4/5} = t^4. Substituting: 01ln(x)x4/5dx=015ln(t)t45t4dt=2501ln(t)dt\int_0^1 \frac{\ln(x)}{x^{4/5}} \, dx = \int_0^1 \frac{5\ln(t)}{t^4} \cdot 5t^4 \, dt = 25 \int_0^1 \ln(t) \, dt

    This integral is divergent because ln(t)\ln(t) \to -\infty as t0t \to 0.

    Conclusion:
    The integral diverges.

Analysis for Part B:

  1. Behavior as xx \to \infty:
    The function ln(x)x4/5\frac{\ln(x)}{x^{4/5}} diminishes as xx \to \infty, since x4/5x^{4/5} dominates ln(x)\ln(x).
    To check convergence, examine: 1ln(x)x4/5dx\int_1^\infty \frac{\ln(x)}{x^{4/5}} \, dx

    Let x=etx = e^t, so ln(x)=t\ln(x) = t and dx=etdtdx = e^t \, dt. Substituting: 1ln(x)x4/5dx=0t(et)4/5etdt=0tet/5dt\int_1^\infty \frac{\ln(x)}{x^{4/5}} \, dx = \int_0^\infty \frac{t}{(e^t)^{4/5}} \cdot e^t \, dt = \int_0^\infty t e^{-t/5} \, dt

    Let u=t/5u = t/5, so t=5ut = 5u and dt=5dudt = 5 \, du. Substituting: 0tet/5dt=05ueu5du=250ueudu\int_0^\infty t e^{-t/5} \, dt = \int_0^\infty 5u e^{-u} \cdot 5 \, du = 25 \int_0^\infty u e^{-u} \, du

    The integral 0ueudu\int_0^\infty u e^{-u} \, du converges (it's a Gamma function, Γ(2)=1\Gamma(2) = 1).
    Hence: 1ln(x)x4/5dx=251=25\int_1^\infty \frac{\ln(x)}{x^{4/5}} \, dx = 25 \cdot 1 = 25

    Conclusion:
    The integral converges, and its exact value is 25.


Final Answers:

  • Part A: Divergent (input 333).
  • Part B: Convergent, exact value is 25.

Would you like a more detailed breakdown of the calculations?


Related Questions:

  1. What substitution techniques are helpful in integrals involving logarithms?
  2. How can you determine divergence at a boundary point?
  3. What is the Gamma function, and how is it used in definite integrals?
  4. How does the rate of growth of logarithmic functions compare to polynomial or exponential functions?
  5. Can the integral in Part A be regularized or reinterpreted to make it finite?

Tip: Always check for singularities at boundaries when analyzing improper integrals.

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Math Problem Analysis

Mathematical Concepts

Improper Integrals
Logarithmic Functions
Substitution Method
Convergence and Divergence

Formulas

Integral of the form ∫ ln(x) / x^(n) dx
Substitution: x = t^5 and x = e^t transformations

Theorems

Convergence of improper integrals at singularities
Gamma function (Γ(n)) for evaluation of definite integrals

Suitable Grade Level

Undergraduate Level